Where are we using the monotone convergence theorem in this proof? I am looking at the proof that $\frac{\sin{x}}{x}$ is not Lebesgue integrable on $[0,\infty)$ from Carrothers. 
(I know that "prove $\frac{\sin{x}}{x}$ is not Lebesgue integrable has been asked before on this site, but the proofs there were provided were different from what Carrothers does, and my question is really more about the MCT than the proof of non-existence, so hopefully this isn't a duplicate! If it is, I apologize).
Carrothers writes: "By the Monotone Convergence Theorem,
$$\int_0^\infty \frac{|\sin{x}|}{x}dx = \sum_{n=1}^\infty \int_{(n-1)\pi}^{n\pi}\frac{|\sin{x}|}{x}dx \ge \sum_{n-1}^\infty \frac{1}{n\pi}\int_0^\pi |\sin(x)|dx = \infty."$$
I'm confused as to where the MCT is used. It seems to me it must be in the first equality, however, I thought that just followed from linearity of the Lebesgue integral? By my reasoning, then, we have
$$\int_{(n-1)\pi}^{n\pi}\frac{|\sin{x}|}{x} = \int_0^\pi\frac{|\sin{x}|}{x+(n-1)\pi}\ge \int_0^\pi\frac{|\sin{x}|}{n\pi},$$
which gives us the inequality, and then the last equality is obvious. So where does the MCT come in?
 A: I bet the MCT is applied in the following way: the sequence of functions given by $$f_N(x)=\frac{\left|\sin x\right|}{x}\cdot\mathbb{1}_{(0,N\pi)}(x)$$
is a sequence of $L^1(\mathbb{R}^+)$ functions that monotonically converge to $f(x)=\frac{\left|\sin x\right|}{x}$ over $\mathbb{R}^+$. Assuming that $f\in L^1(\mathbb{R}^+)$, it follows that:
$$ \int_{0}^{+\infty}f(x)\,dx = \lim_{N\to +\infty}\int_{0}^{+\infty}f_N(x)\,dx =\lim_{N\to +\infty}\sum_{k=0}^{N-1}\int_{0}^{\pi}\frac{\left|\sin x\right|}{x+k\pi}$$
but the last series is a divergent series, hence $f\not\in L^1(\mathbb{R}^+)$.
A: Indeed MCT is uesd in the first step. Linearity only applies to when dealing with finite sums.
Notice that $\left\{[(n-1)\pi,n\pi)\right\}_{n=1}^{\infty}$ is a partition of the positive half-line $[0,\infty)$. Now clearly $$\int_{[0,\infty)}\frac{|\sin(x)|}{x}\,d\lambda(x)= \int_{\bigcup_{n\geq 1}[(n-1)\pi,n\pi)}\frac{|\sin(x)|}{x}\,d\lambda(x)$$
Now we can put the characteristic function inside the integral $$\int_{[0,\infty)}\frac{|\sin(x)|}{x}\,d\lambda(x)=\int_{[0,\infty)}1_{\bigcup_{n\geq 1}[(n-1)\pi,n\pi)}\, \frac{|\sin(x)|}{x} \, d\lambda(x)=\int_{[0,\infty)}\sum_{n=1}^{\infty}1_{[(n-1)\pi,n\pi)}\frac{|\sin(x)|}{x}\, d\lambda(x)$$
Now MCT (Actually Beppo-Levi's) is used on the non-negative, non-decreasing partialsum $$S_{N}(x)=\sum_{n=1}^{N}{1_{[(n-1)\pi,n\pi)}}\frac{|\sin(x)|}{x}$$ which converges pointwise for all $x\in[0,\infty)$ to $$S(x)= \sum_{n=1}^{\infty}1_{[(n-1)\pi,n\pi)}\frac{|\sin(x)|}{x}$$ 
Hence $$\int_{[0,\infty)}S(x) \, d\lambda(x) = \lim_{N\rightarrow \infty}\int_{[0,\infty)}S_{N}(x) \, d\lambda(x)= \lim_{N \rightarrow \infty}\sum_{n=1}^{N}\int_{[(n-1)\pi,n\pi)}\frac{|\sin(x)|}{x} \, d\lambda(x)$$
